Number System Practice Gym by Anubhav Sehgal - Part 2


  • NMIMS, Mumbai (Marketing)


    Q1) Find the sum of all possible values of x such that 2/(sqrt(x)) + 1/(sqrt(y)) = 1/(sqrt(2)) where x and y are positive integers

    u = sqrt(x) and v = sqrt(y)
    2v + u = uv/rt(2)
    uv - 2rt2 v - rt2 u = 0
    v(u - 2rt2) - rt2(u - 2rt2) = 4
    (v - rt2)(u - 2rt2) = 4 = rt2 * 2rt2 = 2rt2 * rt2
    u,v = 4rt2,2rt2 ; 3rt2,3rt2
    x,y = 32,8 ; 18,18
    Sum = 50

    Q2) How many three digit numbers are 34 times of sum of their sum of digits?

    100a + 10b + c = 34a + 34b + 34c
    66a - 24b - 33c = 0
    66a = 24b + 33c
    a = 1 ; b = 0 ; c = 2
    a = 2 ; b = 0 ; c = 4
    a = 3 ; b = 0 ; c = 6
    a = 4 ; b = 0 ; c = 8
    Four such numbers.

    Q3) Find the number of integers, N such that (N^2 + 2N - 8 )/(N^2 + N - 12) is an integer.

    N^2 + 2N - 8 = (N - 2)(N + 4)
    N^2 + N - 12 = (N - 3)(N + 4)
    => (N - 2)/(N - 3) is an integer
    => (N - 3 + 1)/(N - 3) is an integer
    => 1 + 1/(N - 3) is an integer => N = 4. Only one possible value.

    Q4) Find the 1000th term of the sequence : 1,3,4,7,8,9,10,11,13,14,... in which there is no number which contain digit 2,5 or 6.

    I have two approaches for this one :
    The Shareef approach :smile:
    The Tareef approach :stuck_out_tongue_winking_eye:

    The Shareef approach :
    _ : 6 numbers ; _ _ : 6 * 7 = 42 numbers ; _ _ _ : 6 * 7 * 7 = 294 numbers
    342 numbers added till 999
    1 _ _ _ : 7 * 7 * 7 = 343 added = 685 numbers till now. Remaining 1000 - 685 = 315
    3 0 _ _ : 49 added
    3 1 _ _ : 49 added
    3 3 _ _ : 49 added
    3 4 _ _ , 3 7 _ _ , 3 8 _ _ = 49*3 = 147
    Total 294 more. Remaining 21
    3 9 0 _ : 7
    3 9 1 _ : 7
    3 9 3 _ : 7
    3939

    The Tareef approach :
    0 -> 0
    1 -> 1
    2 -> 3
    3 -> 4
    4 -> 7
    5 -> 8
    6 -> 9
    1000th term : 1000 in base 7 = 2626 3939

    Q5) How many positive divisors of 100^10 end in exactly two zeroes?

    100^10 = 2^20 * 5^20
    2^18 * 5^18
    19 + 19 - 1 = 37

    Q6) 960 = 880
    c + 880 + c + 800 + c = 1680 + 3c > 1000 Not to be considered
    OR
    for x c = 1 to 13
    x = -21 to 8
    Option c)

    Q7) How many pairs of non negative integers exist,such that the difference between their product and sum is 72?

    ab - a - b = 72
    ab - a - b + 1 = 73
    a(b - 1) - 1(b - 1) = 73
    (a - 1)(b - 1) = 73 = 1 * 73 = -1 * -73
    a,b = 2,74 ; 74,2 ; 0,-72 ; -72,0
    Non negative integer pairs = 1 unordered or 2 ordered
    Total integral solutions = 2 unordered or 4 ordered
    Option d) Answer : 1

    Q8) The equation 3^(x - 1) + 5^(x - 1) = 34 has :
    a) No solution
    b) 1 solution
    c) 2 solutions
    d) More than 2 solutions

    34 = 25 + 9
    x = 3 is the only solution
    Both are exponentially increasing.
    3 + 5 = 8< 34 ; 9 + 25 = 34 ; 27 + 125 > 34. No other values. Option b)

    Q9) For what smallest positive integer N, will the number 3! + 5! + 7! + .. + (2N + 1)! be a perfect square ?

    3!(1 + 4 * 5 + 4 * 5 * 6 * 7 + ....) = 3! * odd . Power of 2 is 1. So no possible N.

    Q10) What is the unit digit of [10^3000/(10^100 + 3)] where [.] is GIF.

    Remainder = 10^3000/(10^100 + 3)
    Put x = 10^100 => x^30/x + 3. Put x = -3 => (-3)^30 = 3^30
    Dividend = Quotient*Divisor + Remainder
    Check for unit digit on both sides,
    0 = x * 3 + 9
    x = 7


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